**Description:** Every II-finite set (every chain of subsets has a maximal element) is
III-finite (has no denumerable collection of subsets). The proof here
is the only one I could find, from
http://matwbn.icm.edu.pl/ksiazki/fm/fm6/fm619.pdf p.94 (writeup by
Tarski, credited to Kuratowski). Translated into English and modern
notation, the proof proceeds as follows (variables renamed for
uniqueness):

Suppose for a contradiction that A is a set which is II-finite but not III-finite.

For any countable sequence of distinct subsets T of A , we can form a decreasing sequence of nonempty subsets ( UT ) by taking finite intersections of initial segments of T while skipping over any element of T which would cause the intersection to be empty.

By II-finiteness (as fin2i2 ) this sequence contains its intersection, call it Y ; since by induction every subset in the sequence U is nonempty, the intersection must be nonempty.

Suppose that an element X of T has nonempty intersection with Y . Thus, said element has a nonempty intersection with the corresponding element of U , therefore it was used in the construction of U and all further elements of U are subsets of X , thus X contains the Y . That is, all elements of X either contain Y or are disjoint from it.

Since there are only two cases, there must exist an infinite subset of T which uniformly either contain Y or are disjoint from it. In the former case we can create an infinite set by subtracting Y from each element. In either case, call the result Z ; this is an infinite set of subsets of A , each of which is disjoint from Y and contained in the union of T ; the union of Z is strictly contained in the union of T , because only the latter is a superset of the nonempty set Y .

The preceding four steps may be iterated a countable number of times starting from the assumed denumerable set of subsets to produce a denumerable sequence B of the T sets from each stage. Great caution is required to avoid ax-dc here; in particular an effective version of the pigeonhole principle (for aleph-null pigeons and 2 holes) is required. Since a denumerable set of subsets is assumed to exist, we can conclude _om e. _V without the axiom.

This B sequence is strictly decreasing, thus it has no minimum, contradicting the first assumption. (Contributed by Stefan O'Rear, 2-Nov-2014) (Proof shortened by Mario Carneiro, 17-May-2015)

Ref | Expression | ||
---|---|---|---|

Assertion | fin23 | $${\u22a2}{A}\in {\mathrm{Fin}}^{\mathrm{II}}\to {A}\in {\mathrm{Fin}}^{\mathrm{III}}$$ |

Step | Hyp | Ref | Expression |
---|---|---|---|

1 | isf33lem | $${\u22a2}{\mathrm{Fin}}^{\mathrm{III}}=\left\{{g}|\forall {a}\in \left({\mathcal{P}{g}}^{\mathrm{\omega}}\right)\phantom{\rule{.4em}{0ex}}\left(\forall {x}\in \mathrm{\omega}\phantom{\rule{.4em}{0ex}}{a}\left(\mathrm{suc}{x}\right)\subseteq {a}\left({x}\right)\to \bigcap \mathrm{ran}{a}\in \mathrm{ran}{a}\right)\right\}$$ | |

2 | 1 | fin23lem40 | $${\u22a2}{A}\in {\mathrm{Fin}}^{\mathrm{II}}\to {A}\in {\mathrm{Fin}}^{\mathrm{III}}$$ |